Two particles of the same mass are moving in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Since the particles are moving in circular orbits,the centripetal force is provided by the given force $F(r)$.$\frac{mv^2}{r} = |F(r)| = \frac{16}

Vedclass · Accepted Answer

$6 	imes 10^{-2}$

Vedclass · Answer

(B) Since the particles are moving in circular orbits,the centripetal force is provided by the given force $F(r)$.$\frac{mv^2}{r} = |F(r)| = \frac{16}{r} + r^3$Multiplying both sides by $\frac{r}{2}$,we get the kinetic energy $K = \frac{1}{2}mv^2 = \frac{1}{2}(16 + r^4)$.For the first particle at $r = 1$:$K_1 = \frac{1}{2}(16 + 1^4) = \frac{17}{2} = 8.5$.For the second particle at $r = 4$:$K_2 = \frac{1}{2}(16 + 4^4) = \frac{1}{2}(16 + 256) = \frac{272}{2} = 136$.The ratio of kinetic energies is:$\frac{K_1}{K_2} = \frac{17/2}{272/2} = \frac{17}{272} = \frac{1}{16} = 0.0625$.Thus,the ratio is approximately $6 	imes 10^{-2}$.

Two particles of the same mass are moving in circular orbits because of a force given by $F(r) = -\frac{16}{r} - r^3$. The first particle is at a distance $r = 1$,and the second is at $r = 4$. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to:

Similar Questions

The tube $AC$ forms a quarter circle in a vertical plane. The ball $B$ has an area of cross-section slightly smaller than that of the tube,and can move without friction through it. $B$ is placed at $A$ and displaced slightly. It will

$A$ motorcycle is moving on an overbridge of radius $R$. The driver maintains a constant speed $v$. As the motorcycle is ascending on the overbridge,the normal force on it is

$A$ coin,placed on a rotating turntable,slips when it is placed at a distance of $9 \, cm$ from the center. If the angular velocity of the turntable is tripled,it will just slip if its distance from the center is ......... $cm$.

$A$ particle of mass $m$ is moving in a circular path of radius $r$ in a plane. Its angular momentum is $L$. The value of the centripetal force acting on the particle is:

$A$ particle is moving with constant speed along a uniform helical path $(a)$ and a spiral path $(b)$ separately,as shown in the figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module